# Discrete Random Variables Probability Exercise - How to approach this

Below is the whole exercise that I need to solve.

Since this is from an online course and it's given with any other context, I need to figure what I need to learn in order to solve it.

Is it Binomial Distribution?

Any approach to the way to solve each of them would be awesome, because it would allow me to find a book and study the relevant part.

Let $X$ and $Y$ be $2$ discrete random variables which probability set function is defined by:

• $f(1,3)=0.1$
• $f(1,5)=0.3$
• $f(2,3)=0.4$
• $f(2,5)=0.2$

1. Without determining the marginal probability function for $X$ and $Y$, please solve, justifying your answers:

• The value of $P(X=1,Y=<4)$ and $P(X=2|Y=3)$

• The value of $E(X)$ and of $E(Y)$

• The covariance between $X$ and $Y$

2. Find the marginal distribution function of $X$

-
You have four separate cases, each with a specified probability of happening. For the first question, find all cases where $X=1$ and $Y\leq 4$, and add up their probabilities. Use the definitions of conditional probability, expectation value, etc. to answer the other questions. – user3294068 May 22 '14 at 20:47

$P(A|B) = \begin{cases} \frac{P(A \cap B)}{P(B)} & \text{$P(B)>0$}\\ 0 & \text{$P(B)=0$}\\ \end{cases}$

Therefore:

• $P(X=2|Y=3) = \frac{P(X=2 \cap Y=3)}{P(Y=3)} = \frac{0.4}{0.1+0.4} = 0.8$

$E(N) = \sum\limits_{i=1}^{k}N_iP_i = N_1P_1+N_2P_2+\dots+N_kP_k$

Therefore:

• $E(X) = 1\cdot0.1+1\cdot0.3+2\cdot0.4+2\cdot0.2 = 1.6$
• $E(Y) = 3\cdot0.1+5\cdot0.3+3\cdot0.4+5\cdot0.2 = 4.0$
-
Thank you! Whats the name of the formula for the expected value of N so I can look it up. Sorry if the question sounds terrible, but I approaching the whole probability subject without any base knowledge – peppp May 22 '14 at 22:25
@peppp: You're welcome. en.wikipedia.org/wiki/Conditional_probability for the first formula, en.wikipedia.org/wiki/Expected_value for the second formula. – barak manos May 22 '14 at 22:30